Question: If $x \oplus y = x^{2}-3y^{2}$ and $x \diamond y = 6x-4y$, find $0 \diamond (-6 \oplus -3)$.
Answer: First, find $-6 \oplus -3$ $ -6 \oplus -3 = (-6)^{2}-3(-3)^{2}$ $ \hphantom{-6 \oplus -3} = 9$ Now, find $0 \diamond 9$ $ 0 \diamond 9 = (6)(0)-(4)(9)$ $ \hphantom{0 \diamond 9} = -36$.